The exterior calculus discussed previously, including the integration of differential forms that will be briefly touched upon later, consists of purely algebraic definitions and does not inherently possess any geometric meaning. However, we can associate certain formulas or definitions with geometric content to deepen our understanding and apply them more flexibly. Nevertheless, this is merely a correspondence and depends on our interpretation. For example, we say that the exterior derivative formula \int_{\partial D} Pdx+Qdy = \int_{D} \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\land dy \tag{32} corresponds to Green’s Theorem \int_{\partial D} Pdx+Qdy = \int_{D} \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy \tag{33} This is perfectly fine, but they are not equivalent; they just happen to be identical in form. Green’s Theorem describes the relationship between a line integral over a closed curve and a surface integral, whereas the exterior derivative formula is a purely algebraic operation. You could just as easily map dx\land dy to -dxdy instead of dxdy, which would result in a different geometric correspondence.
A more profound question is: why does this correspondence exist? That is to say, why can we obtain a correspondence with integral formulas after some adjustments and interpretations? First, it must be clarified that the exterior product, aside from its antisymmetry, is no different from the product of ordinary numbers; thus, many properties are preserved. Secondly, we should return to the concept of antisymmetry itself: the determinant of a matrix represents the volume of the n-dimensional solid spanned by the vectors corresponding to the matrix. Since the determinant is antisymmetric, this implies that antisymmetric operations have an inherent connection with volume and integration. Of course, I have yet to achieve a more detailed understanding of this.
Furthermore, when we seek the geometric meaning of differential forms, we usually discuss spaces of no more than three dimensions. It is difficult to imagine geometric images in higher dimensions, especially for high-dimensional surface integrals; these are generally just analogies, and whether the analogy holds sometimes requires further deliberation. Therefore, in such cases, it is better to simply say that what differential forms describe is geometry itself, rather than searching for a so-called geometric meaning. In other words, one can reverse the perspective and define geometry using differential forms and exterior derivatives as axiomatic first principles.
One could even treat exterior calculus merely as an effective way to memorize various differential and integral formulas. For instance, if I were to ask everyone to write down Stokes’ Theorem in three-dimensional space from memory, many might get confused about which term subtracts from which. However, within the framework of exterior calculus, it can be derived very quickly. It is like Equation (11); even if one does not seek a geometric explanation, it represents Kepler’s Second Law: equal areas are swept out in equal time. Yet, even without the geometric explanation, you can still solve the equation.
The following sections will expand on geometric interpretations.
Exterior Product: Spanning and Projecting
Consider the exterior product of two differential 1-forms, for example: \begin{aligned} \alpha_{\mu}dx^{\mu} \land \beta_{\nu}dx^{\nu} &= \alpha_{\mu}\beta_{\nu} dx^{\mu}\land dx^{\nu}\\ &= \sum_{\mu < \nu} (\alpha_{\mu}\beta_{\nu}-\beta_{\mu}\alpha_{\nu}) dx^{\mu}\land dx^{\nu} \\ &= \frac{1}{2}(\alpha_{\mu}\beta_{\nu}-\beta_{\mu}\alpha_{\nu}) dx^{\mu}\land dx^{\nu} \end{aligned} \tag{34} When the summation symbol is omitted, it indicates that \mu and \nu each traverse their range without constraints. Note that \alpha_{\mu}\beta_{\nu}-\beta_{\mu}\alpha_{\nu}=\det\begin{pmatrix}\alpha_{\mu}&\alpha_{\nu}\\\beta_{\mu}&\beta_{\nu}\end{pmatrix} \tag{35} If we view dx^{\mu} as a basis, then for a selected pair \mu, \nu, the term \alpha_{\mu}\beta_{\nu}-\beta_{\mu}\alpha_{\nu} corresponds exactly to the oriented area of the projection of the parallelogram spanned by vectors \alpha and \beta onto the dx^{\mu}, dx^{\nu} plane.
For general differential p-forms and q-forms, their exterior product can be constructed similarly, though higher dimensions are harder to visualize. For example, the exterior product of a 1-form and a 2-form can be imagined as a vector and an "area vector" (actually a tensor) spanning a parallelepiped. Each term of the exterior product result is the volume of the projection of that parallelepiped onto the corresponding three-dimensional subspace, and so on. Specifically, in an n-dimensional space, if n differential 1-forms are multiplied, the result is: \alpha_{\mu_1}^{1} dx^{\mu_1}\land \dots \land \alpha_{\mu_n}^{n} dx^{\mu_n}=\det(\alpha_{\mu}^{\nu}) dx^1 \land \dots \land dx^n \tag{36} This produces the determinant of a matrix, which is quite remarkable and is a direct manifestation of antisymmetry. Antisymmetry also exists in determinants: swapping two rows or two columns changes the sign of the determinant. Let f be an arbitrary function; we have df=\frac{\partial f}{\partial x^{\mu}} dx^{\mu}, then df^1 \land \dots \land df^n = \det\left(\frac{\partial f^{\mu}}{\partial x^{\nu}}\right) dx^1 \land \dots \land dx^n \tag{37} From the perspective of transformation, \det\left(\frac{\partial f^{\mu}}{\partial x^{\nu}}\right) is the Jacobian determinant of the integral transformation. This tempts us to ignore the \land symbol and treat dx^1 \land \dots \land dx^n directly as the volume element dx^1 \dots dx^n. In fact, this is exactly what is done! We define dx^{\mu_1}\land \dots \land dx^{\mu_k}=\pm dx^{\mu_1}\dots dx^{\mu_k}, where the sign depends on the specific geometric content we wish to interpret. In this way, we can use exterior calculus to represent integral theory.
Differential Operator: Going in Circles
What is even more worthy of profound understanding is Equation (25), which describes what happens when moving from a p-form to a p+1 form, or what geometric content it corresponds to.
Let us start with a differential 1-form \omega_{\nu} dx^{\nu}. Under the action of the operator d, we have: \begin{aligned} d(\omega_{\nu} dx^{\nu}) &= \frac{\partial \omega_{\nu}}{\partial x^{\mu}} dx^{\mu} \land dx^{\nu}\\ &= \sum_{\mu < \nu} \left(\frac{\partial \omega_{\nu}}{\partial x^{\mu}}-\frac{\partial \omega_{\mu}}{\partial x^{\nu}}\right) dx^{\mu} \land dx^{\nu}\\ &= \frac{1}{2} \left(\frac{\partial \omega_{\nu}}{\partial x^{\mu}}-\frac{\partial \omega_{\mu}}{\partial x^{\nu}}\right) dx^{\mu} \land dx^{\nu} \end{aligned} \tag{38} What is the geometric correspondence for something of this shape? We can view \omega_{\nu} dx^{\nu} as the increment of a quantity \Omega from x to x+dx, i.e., \Omega (x+dx) = \Omega (x) + \omega_{\nu}(x) dx^{\nu} \tag{39} Then, if we move from x+dx to x+dx+\delta x, we naturally have: \begin{aligned} \Omega_1 (x+dx+\delta x) &= \Omega (x+dx) + \omega_{\nu}(x+dx) \delta x^{\nu}\\ &= \Omega (x) + \omega_{\nu}(x) dx^{\nu} + \omega_{\nu}(x) \delta x^{\nu} + \frac{\partial \omega_{\nu}}{\partial x^{\mu}} dx^{\mu} \delta x^{\nu} \end{aligned} \tag{40} This follows the path x \to x+dx \to x+dx+\delta x. Swapping dx and \delta x, i.e., following the path x \to x+\delta x \to x+\delta x+ dx, yields: \Omega_2 (x+dx+\delta x) = \Omega (x) + \omega_{\nu}(x) \delta x^{\nu} + \omega_{\nu}(x) d x^{\nu} + \frac{\partial \omega_{\nu}}{\partial x^{\mu}} \delta x^{\mu} d x^{\nu} \tag{41} The difference between the two: \begin{aligned} \left(\frac{\partial \omega_{\nu}}{\partial x^{\mu}}-\frac{\partial \omega_{\mu}}{\partial x^{\nu}}\right)dx^{\mu} \delta x^{\nu} &= \frac{1}{2}\left(\frac{\partial \omega_{\nu}}{\partial x^{\mu}}-\frac{\partial \omega_{\mu}}{\partial x^{\nu}}\right)(dx^{\mu} \delta x^{\nu}-dx^{\nu} \delta x^{\mu})\\ &= \frac{1}{2}\left(\frac{\partial \omega_{\nu}}{\partial x^{\mu}}-\frac{\partial \omega_{\mu}}{\partial x^{\nu}}\right) \det\begin{pmatrix}dx^{\mu} & \delta x^{\mu}\\ dx^{\nu} & \delta x^{\nu}\end{pmatrix} \end{aligned} \tag{42} is the amount of change produced after wandering around the closed path x \to x+dx \to x+dx+\delta x \to x+\delta x \to x.
If we let dx^{\mu} \land dx^{\nu} = \det\begin{pmatrix}dx^{\mu} & \delta x^{\mu}\\ dx^{\nu} & \delta x^{\nu}\end{pmatrix} \tag{43} then Equation (42) is exactly d(\omega_{\nu} dx^{\nu}). And \det\begin{pmatrix}dx^{\mu} & \delta x^{\mu}\\ dx^{\nu} & \delta x^{\nu}\end{pmatrix} is the area of the projection of the parallelogram spanned by the two vectors dx and \delta x onto the x^{\mu}, x^{\nu} plane, which is also antisymmetric. From this perspective, dx^{\mu} \land dx^{\nu} can be interpreted as an oriented area element, and the meaning of d(\omega_{\nu} dx^{\nu}) is the change in a quantity after traversing a small loop!
The Fundamental Theorem of Calculus
Through the "looping" approach, we explained the meaning of moving from a 1-form to a 2-form. Unfortunately, moving from a general p-form to a p+1 form is not as easy to visualize, and in fact, it is difficult to imagine the geometric image of integration in spaces exceeding three dimensions. Therefore, we use a path that "reverses the logical order." If \omega is a differential p-form and D is a given region, then: \int_{\partial D} \omega = \int_{D} d\omega \tag{44} That is, the integral of \omega over the boundary is equal to the integral of d\omega over the region. This is the "Stokes’ Theorem" in the context of differential forms, or one could say it is the Fundamental Theorem of Calculus in exterior calculus.
The celebrated aspect of this formula is that it unifies the Newton-Leibniz formula, Green’s Theorem, Gauss’s Theorem, and Stokes’ Theorem, and generalizes them. One might be confused by "what is the integration of a differential form?" In fact, there is nothing special about the integration of differential forms, because expressions like dx^{\mu}\land dx^{\nu}, aside from being antisymmetric, are no different from ordinary differential elements dx^{\mu}dx^{\nu}, and the definition of an integral (such as using the Riemann integral definition) is independent of symmetry or antisymmetry.
Thus, we can imagine that moving from a general p-form to a p+1 form, or from \omega to d\omega, is essentially the same as moving from a 1-form to a 2-form—it involves a similar "looping" process. That is, \omega is understood as the change along the boundary, while d\omega is the change produced after traversing a small region. It then follows naturally that for a closed region D, we have \int_{\partial D} \omega = \int_{D} d\omega.
Of course, as mentioned before, this is a "logical reversal." This integral theorem is actually a "result" rather than a "cause"; it requires a lengthy proof. We have cited it here without proof to provide a path that is as quick and clear as possible to explain the meaning of d\omega.
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